Lithium ii sulfide is the answer to your question
Answer:
K+
Explanation:
Potassium(an alkali metal) belongs to element in group 1 and period 4 in the periodic table. It has to release one electron to attain noble configuration.
Percentage Weight-in-volume is defined as the <em><u>number of grams of a solute in a 100 ml (milliliters) solution.</u></em>
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<u>Percentage Weight-in-volume</u> can tell us about the <em>degree of concentration of a given solution.</em>
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The solute can be <em>crystalline or non-crystalline in nature.</em>
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The <u>number of grams of glucose</u> present in a <u>5% glucose solution</u> is 5 grams.
- This question is based on a Percentage Weight-in-volume. The formula states that:
a% of a glucose solution =<u> a grams of glucose in a 100 mL solution</u>
Hence, 5% glucose solution = 5 grams of glucose / 100 mL solution
Therefore, the <u>number of grams of glucose</u> present in a <u>5% glucose solution</u> is 5 grams.
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Answer:
[H₃O⁺] = [F⁻] = 2.2 x 10⁻² M. & [OH⁻] = 4.55 x 10⁻¹³.
Explanation:
- For a weak acid like HF, the dissociation of HF will be:
<em>HF + H₂O ⇄ H₃O⁺ + F⁻.</em>
[H₃O⁺] = [F⁻].
<em>∵ [H₃O⁺] = √Ka.C,</em>
Ka = 6.8 x 10⁻⁴, C = 0.710 M.
∴ [H₃O⁺] = √Ka.C = √(6.8 x 10⁻⁴)(0.710) = 2.197 x 10⁻² M ≅ 2.2 x 10⁻² M.
<em>∴ [H₃O⁺] = [F⁻] = 2.2 x 10⁻² M.</em>
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∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.
<em>∴ [OH⁻] = 10⁻¹⁴/[H₃O⁺]</em> = 10⁻¹⁴/(2.2 x 10⁻²) = <em>4.55 x 10⁻¹³.</em>